Solve 4x^2+8x-45

$$4 x ^ { 2 } + 8 x - 45$$

Factor

$$\left(2x-5\right)\left(2x+9\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx-45. To find a and b, set up a system to be solved.

$$a+b=8 ab=4\left(-45\right)=-180$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 8.

$$a=-10 b=18$$

Rewrite 4x^{2}+8x-45 as \left(4x^{2}-10x\right)+\left(18x-45\right).

$$\left(4x^{2}-10x\right)+\left(18x-45\right)$$

Factor out 2x in the first and 9 in the second group.

$$2x\left(2x-5\right)+9\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(2x+9\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4x^{2}+8x-45=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-8±\sqrt{8^{2}-4\times 4\left(-45\right)}}{2\times 4}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 4\left(-45\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-8±\sqrt{64-16\left(-45\right)}}{2\times 4}$$

Multiply -16 times -45.

$$x=\frac{-8±\sqrt{64+720}}{2\times 4}$$

Add 64 to 720.

$$x=\frac{-8±\sqrt{784}}{2\times 4}$$

Take the square root of 784.

$$x=\frac{-8±28}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-8±28}{8}$$

Now solve the equation x=\frac{-8±28}{8} when ± is plus. Add -8 to 28.

$$x=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{-8±28}{8} when ± is minus. Subtract 28 from -8.

$$x=-\frac{36}{8}$$

Reduce the fraction \frac{-36}{8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{9}{2}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5}{2} for x_{1} and -\frac{9}{2} for x_{2}.

$$4x^{2}+8x-45=4\left(x-\frac{5}{2}\right)\left(x-\left(-\frac{9}{2}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$4x^{2}+8x-45=4\left(x-\frac{5}{2}\right)\left(x+\frac{9}{2}\right)$$

Subtract \frac{5}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+8x-45=4\times \frac{2x-5}{2}\left(x+\frac{9}{2}\right)$$

Add \frac{9}{2} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+8x-45=4\times \frac{2x-5}{2}\times \frac{2x+9}{2}$$

Multiply \frac{2x-5}{2} times \frac{2x+9}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$4x^{2}+8x-45=4\times \frac{\left(2x-5\right)\left(2x+9\right)}{2\times 2}$$

Multiply 2 times 2.

$$4x^{2}+8x-45=4\times \frac{\left(2x-5\right)\left(2x+9\right)}{4}$$

Cancel out 4, the greatest common factor in 4 and 4.

$$4x^{2}+8x-45=\left(2x-5\right)\left(2x+9\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 4

$$x ^ 2 +2x -\frac{45}{4} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -2 rs = -\frac{45}{4}$$

Two numbers r and s sum up to -2 exactly when the average of the two numbers is \frac{1}{2}*-2 = -1. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -1 - u s = -1 + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{45}{4}

$$(-1 - u) (-1 + u) = -\frac{45}{4}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$1 - u^2 = -\frac{45}{4}$$

Simplify the expression by subtracting 1 on both sides

$$-u^2 = -\frac{45}{4}-1 = -\frac{49}{4}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{49}{4} u = \pm\sqrt{\frac{49}{4}} = \pm \frac{7}{2}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-1 - \frac{7}{2} = -4.500 s = -1 + \frac{7}{2} = 2.500$$

Evaluate

$$\left(2x-5\right)\left(2x+9\right)$$